Geodesics and stretched strings

Consider this real situation: Light from a far-background galaxy is observed to be lensed by a closer massive galaxy. Individual photons that graze the closer galaxy travel to the observer along various geodesic paths, as prescibed by GR.

Now consider a very hypothetical situation. Suppose a point on the background galaxy is connected to the observer by a weightless stretched string. Does this string lie along a geodesic?

Consider this real situation: Light from a far-background galaxy is observed to be lensed by a closer massive galaxy. Individual photons that graze the closer galaxy travel to the observer along various geodesic paths, as prescibed by GR.

Now consider a very hypothetical situation. Suppose a point on the background galaxy is connected to the observer by a weightless stretched string. Does this string lie along a geodesic?

A (timelike or lightlike) geodesic is a 1-dimensional worldline in spacetime. A string traces out a 2-dimensional worldsheet in spacetime.

I am going to try and formulate two versions of your question. You might have something totally different in mind, and I don't really know the answer to either version.

Version 1: If a magic marker is used to put a small red dot on the string at an point on the string, is the worldline of the dot a geodesic in spacetime?

If there are (non-uniform?) stresses in the string, probably not.

On to the second version.

Define space as a 3-dimensional hypersurface in spacetime given by t = constant, where t is some appropriate timelike (i.e., [itex]\partial / \partial t[/itex] is a future-directed timelike vector field) coordinate. In other words, space is a hypersurface of "simultaneity".

The intersection of the string's worldsheet with space is a 1-dimensional curve in space. Restricting the spacetime metric to space naturally defines a positive-definite 3-metric on space.

Version 2: Is the string's 1-dimensional curve in space a geodesic in space with respect to this spatial metric?

A (timelike or lightlike) geodesic is a 1-dimensional worldline in spacetime. A string traces out a 2-dimensional worldsheet in spacetime.

I am going to try and formulate two versions of your question. .....
.... the second version.

Define space as a 3-dimensional hypersurface in spacetime given by t = constant, where t is some appropriate timelike (i.e., [itex]\partial / \partial t[/itex] is a future-directed timelike vector field) coordinate. In other words, space is a hypersurface of "simultaneity".

The intersection of the string's worldsheet with space is a 1-dimensional curve in space. Restricting the spacetime metric to space naturally defines a positive-definite 3-metric on space.

Version 2: Is the string's 1-dimensional curve in space a geodesic in space with respect to this spatial metric?

Thanks for recasting my naive question so clearly. Version 2 is the one I'd like answered..... I think. The answer may well be "yes" , but I really don't know, since the "1-dimensional curve" of the stretched string is shaped ultimately by interactions that are not gravitational (strong and electroweak).

I think the answer is almost always no. Consider e.g. a planet in a circular orbit around its star. The orbit is a geodesic in space-time, but not in space.

You are correct about a planet, which is a body freely falling along a spacetime geodesic. But the stretched string I'm asking about is not a body in free fall, so the answer you give is inapplicable. But thanks for the interest!

Perhaps one should think of a static situation, like the Schwarzchild geometry. Consider first this geometry, with no central mass, where the geodesics followed by light rays and occupied by stretched strings are what we call "straight lines" in space sections.

Now turn on the central mass, quasi-statically, as if you were slowly creating the sun. Consider a grazing light ray, say from a distant star, as in the 1918 expedition. This light becomes perceptibly deviated as you turn the sun on. Would a string stretched from the star to Eddington also become bent along a grazing geodesic? Or not?

the stretched string I'm asking about is not a body in free fall, so the answer you give is inapplicable.

You're right. I didn't read enough of your post and George's reply to understand the question. How about this then? Imagine two very tall buildings some distance from each other. Tie one end of the string to the top of one of them and the other end to the top of the other. The string will have a curved shape. If we increase the string tension (i.e. make the string shorter), the shape will be less curved. How "stretched" do you want your string to be? Are we talking about the limit where string tension goes to infinity? In that case, it seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, but I suppose we'd have to do some calculations to know for sure.

... Now turn on the central mass, quasi-statically, as if you were slowly creating the sun. Consider a grazing light ray, say from a distant star, as in the 1918 expedition. This light becomes perceptibly deviated as you turn the sun on. Would a string stretched from the star to Eddington also become bent along a grazing geodesic? Or not?

I think even your unphysical 'massless, stretched string' will sag towards the Sun in the gravitational field, unless it consists out of massless particles moving along the 'string' at c, in which case it is the same as a grazing light beam!

You're right. I didn't read enough of your post and George's reply to understand the question. How about this then? Imagine two very tall buildings some distance from each other. Tie one end of the string to the top of one of them and the other end to the top of the other. The string will have a curved shape. If we increase the string tension (i.e. make the string shorter), the shape will be less curved. How "stretched" do you want your string to be? Are we talking about the limit where string tension goes to infinity? In that case, it seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, but I suppose we'd have to do some calculations to know for sure.

In a strong gravitational field a light beam shone from one building to the other will follow a curved path a bit like the trajectory of a cannon ball rising at first and then falling ( a parabola?). The string on the other hand starts with a sag and tends towards a straight line as the tension tends towards infinity, so it would seem the string with extreme tension would not approach the path of the light no matter how much tension is applied to it. In other words a string under tension does not follow a null geodesic which is the answer to the OP of this thread IMHO. I am assuming of course that the tension does not cause an antigravity effect that causes it to curve upward, but that seems unlikely.

In another thread https://www.physicsforums.com/showthread.php?t=238428 (post #14 onwards) there is currently an argument as to why light curves in a gravitational field. Is it because the light has inertial and gravitational mass due to its energy, or is it because gravity is simply a geometrical effect that causes object to follow paths determined completely by their velocity irrespective of whether they have any form of mass or not? The answer to the question in that thread reflects on the answer to the question in this thread.

Another approach is this. Light passing from a vacuum to a medium with a refractive index not equal to unity will follow a bent path that minimises the time for light to get from one point to another. On the other hand, a string under tension follows a path that minimises the distance from one point to another. This point of view suggests that path of string under tension and the path of a photon is not the same.

......seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, .

I was thinking of a massless string. I think the answer is this: a light ray grazing the sun traces a geodesic path though spacetime which is not a "straight line" in the space section, near the sun. A massless stretched string would also lie along this geodesic, which is as straight a line as you could hope to define in the curved part of the space section. But maybe I'm wrong?

I think even your unphysical 'massless, stretched string' will sag towards the Sun in the gravitational field, unless it consists out of massless particles moving along the 'string' at c, in which case it is the same as a grazing light beam!

Jorrie

It might melt, but won't sag! On reflection I think that in any case it would appear to lie along the grazing light beam, just like a stick appears bent when poked into water. Dankie, Jorrie.

If the solar wind and other such phenomena are left out of the equation, how can any physical line not sag under the gravitational acceleration? Consider a Schwarzschild black hole with the string passing the hole at r=3M, where light can orbit. If there were something like a static massless particle from which your string could be made, those particles would have to be forced out of their spacetime geodesics (at closest approach) by a radial force component, which you do not have in a straight string - you need some sag for that!

......how can any physical line not sag under the gravitational acceleration?

First, I was considering only an (imaginary) massless string, not a physical line. No mass, no gravitational acceleration! Second, if a physical string is strong enough, and stretched tightly enough, the sag needed to balance gravitational forces can, in principle at least, be made as small as you like. A massless string seems a reasonable model in a practical case, such as that of sun-grazing rays, where melting seems the greater hazard. I agree that in the case of a physical string grazing a black hole at r = 3M sags might be big, but you perhaps have stellar-massed black holes in mind? What about really massive black holes (10^ 6 to 9 solar masses)? Aren't the forces then smaller at r =3M?

It might melt, but won't sag! On reflection I think that in any case it would appear to lie along the grazing light beam, just like a stick appears bent when poked into water. Dankie, Jorrie.

The apparent bent path of a straight stick poked into water does not follow the bent path of the light rays from the stick.

Have a look at the attached diagram. The red line is the straight stick lying on the physical path ABC. The blue line is the apparent position of the stick lying on the path DBC. The average path followed by the light rays is on the path AEC.

Substitute a gravitational field for the water and the path of the photons is AEC and the path of the massless string unaffected by gravity (or water) is along ABC and the apparent path of the string is along DBC. So the answer to your OP is no.

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Here is another diagram. It shows the geodesics of a low velocity particle, a light ray, a low tension string with mass and a high tension string with mass compared to the path of a massless string. In a gravitational field none of the paths lie on the path of the massless string. Switch off the gravitational field and all the paths follow the path of the massless string.

I think I was pretty much "on the money" when I stated:

1) A particle with motion follows a path that minimises the time between two points.

2) An object with tension follows a path that minimises the distance between two points.

A particle with motion follows a path that tends towards a straight line as the velocity tends towards to infinity but it is limited by the speed of light, so it can never follow a straight horizontal path in a gravitational field.

An object with tension follows a path that tends towards a straight line as the tension tends towards infinity but presumably no real material can have infinite strength so this path is never truely a straight horizontal line in a gravitational field.

Only an object with no motion and no mass (a hypothetical massless string) can follow a true horizontal straight line in a gravitational field.

In the absense of readily available massless string in the hardware stores, if you want to build a perfectly straight garden wall you will have to take the average of the builders chalk line and laser leveling device.

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First, I was considering only an (imaginary) massless string, not a physical line. No mass, no gravitational acceleration!

I'm not convinced. It is surely true in Newtonian mechanics, but I can't agree for relativistic mechanics. Spacetime geodesics (even null-geodesics) don't care about the proper mass of entities. They all 'fall' in the same way, AFAIK.

I agree that in the case of a physical string grazing a black hole at r = 3M sags might be big, but you perhaps have stellar-massed black holes in mind? What about really massive black holes (10^ 6 to 9 solar masses)? Aren't the forces then smaller at r =3M?

Yes, the radial gravitational acceleration at r=3M decreases linearly with M, but only approaches zero as M => infinity. In such a case the acceleration will approach zero, because the event horizon circumference will approach infinite length. In all other cases, the radial acceleration will be non-zero.

The apparent bent path of a straight stick poked into water does not follow the bent path of the light rays from the stick.

Have a look at the attached diagram. The red line is the straight stick lying on the physical path ABC. The blue line is the apparent position of the stick lying on the path DBC. The average path followed by the light rays is on the path AEC.

Substitute a gravitational field for the water and the path of the photons is AEC and the path of the massless string unaffected by gravity (or water) is along ABC and the apparent path of the string is along DBC. So the answer to your OP is no.

Yes, you're quite right. Nice diagram -- thanks for all the trouble you've taken. And I agree with your "no", as being the answer to both my posts, 1 and 5. I was too hasty in drawing conclusions from the similarities between the situation I was considering in say post #5 to that of a stick in water.

Your second post seems to address some points raised by Jorrie rather than the questions of my posts 1 and 5, so I'll give it a bit more thought before replying more fully.

Yes, you're quite right. Nice diagram -- thanks for all the trouble you've taken. And I agree with your "no", as being the answer to both my posts, 1 and 5. I was too hasty in drawing conclusions from the similarities between the situation I was considering in say post #5 to that of a stick in water.

Your second post seems to address some points raised by Jorrie rather than the questions of my posts 1 and 5, so I'll give it a bit more thought before replying more fully.

Thanks ,I was just going to post a link to a googled image but believe it or not I couldn't find one which is surprising for such a basic often quoted optical illusion!

A new question also occured to me. How do we know that the apparent bending of a rod poked into water is an optical illusion and that the rod does not "really" bend and repair itself when withdrawn. The answer is that we could measure the stress on the rod and note that the sides of the rod do not experience tension or compression on its sides. This may give an method to define a straight line in a gravitational field. Start with a straight rod far away from a gravitational field and when the rod is lowered into a gravitational field support it in such away that it does not experience any stresses. The supported stress free rod would aproximate the hypothetical massless string that you suggested.

Yes, I agree, but the questions I asked in posts #1 and #5 weren't about the sagging of strings at all. Kev has given an interesting and (I now think after making mistakes) correct answer. This is that starlight grazing the sun follows path that is different from string imagined to be somehow stretched between the star and an observer here on earth.

This seems true whether you imagine the string to have mass and sag, or be massless and not sag. It's the difference that is significant, not how we choose to imagine what the string is made of, how tightly it's stretched or how to tie it to the star! I'll discuss this difference after some more thought, in another post.

Thanks for the clarification of how gravitational acceleration ar r = 3M varies with M. I suppose tidal forces vary faster than linearly?

Thanks for the clarification of how gravitational acceleration ar r = 3M varies with M. I suppose tidal forces vary faster than linearly?

Yep, where gravitational acceleration at r=3M is inversely proportional to the mass of the hole, tidal gravity is approximately proportional to the inverse square of the mass. It is roughly proportional to the inverse third power of r, which in itself is directly proportional to mass; hence the inverse square law.